Question #5d572

Question

2791 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-26T07:23:29

Using the definitions of #csc# and #cot#, along with the identities

we have

#csc(2x)+cot(2x) = 1/sin(2x)+cos(2x)/sin(2x)#

#=(1+cos(2x))/sin(2x)#

#=(1+(2cos^2(x)-1))/(2sin(x)cos(x))#

#=(2cos^2(x))/(2sin(x)cos(x))#

#=cos(x)/sin(x)#

#=cot(x)#

Answer 2 · 2016-12-26T07:40:12

See proof below

We use

#cscx=1/sinx#

#sin2x=2sinxcosx#

#cos2x=2cos^2x-1#

#cotx=cosx/sinx#

So,

#csc2x+cot2x=1/(sin2x)+(cos2x)/(sin2x)#

#=(1+cos2x)/(sin2x)#

#=(1+cos^2x-1)/(2sinxcosx)#

#=(2cos^2x)/(2sinxcosx)#

#=cosx/sinx#

#=cotx#

Q.E.D